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AP Precalculus -2.14 Logarithmic Function Modeling- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -2.14 Logarithmic Function Modeling- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -2.14 Logarithmic Function Modeling- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question 

The table gives ordered pairs \((x, \ln y)\). For the function \(y = f(x)\), which of the following statements about \(f\) is supported by the data in the table?
\(x\)5678
\(\ln y\)36912
(A) The function \(f\) is logarithmic because the values of \(x\) and the values of \(\ln y\) both form arithmetic sequences.
(B) The function \(f\) is linear because the values in each column form an arithmetic sequence.
(C) The function \(f\) is exponential because the values of \(x\) and the values of \(\ln y\) both form arithmetic sequences.
(D) The function \(f\) is exponential because the values of \(\ln y\) increase faster than the values of \(x\).
▶️ Answer/Explanation
Detailed solution

Both \(x\) and \(\ln y\) increase by a constant difference: \(x\) increases by 1, and \(\ln y\) increases by 3. This means \(\ln y\) is a linear function of \(x\). Let \(\ln y = mx + b\). Using points (5,3) and (6,6), slope \(m = 3\), and equation \(\ln y = 3x – 12\). Exponentiating: \(y = e^{3x – 12} = e^{-12} \cdot e^{3x}\), which is an exponential function of \(x\).
✅ Answer: (C)

Question 

Consider the function \(f\). The table gives values of \(\log_3(f(x))\) for selected values of \(x\). Which of the following is a graph of \(y = f(x)\)?
\(x\)0123
\(\log_3(f(x))\)2345

▶️ Answer/Explanation
Detailed solution

From the table, \(\log_3(f(x))\) increases by 1 as \(x\) increases by 1, so \(\log_3(f(x))\) is linear: \(\log_3(f(x)) = x + 2\).
Solve for \(f(x)\): \(f(x) = 3^{x+2} = 3^x \cdot 3^2 = 9 \cdot 3^x\).
This is an exponential function with base 3 and initial value 9 when \(x = 0\). Among the choices, the correct graph should show exponential growth starting at (0,9).
✅ Answer: (D)

Question 

The table gives the weight \(y\), in pounds, of an animal for selected ages \(x\), in years. A logarithmic regression is used to model these data. What is the weight of the animal, to the nearest pound, predicted by the logarithmic function model at age \(4.5\) years?
\(x\) (years)1.522.5346
\(y\) (pounds)242031503615410548356355
(A) 5073
(B) 5203
(C) 5333
(D) 5345
▶️ Answer/Explanation
Detailed solution

1. Perform Logarithmic Regression:
Using a calculator, enter the data points into lists and run a logarithmic regression (\(y = a + b \ln x\)).
The resulting equation is approximately \(y \approx 1167.06 + 2777.95 \ln x\).

2. Predict for \(x = 4.5\):
Substitute \(x = 4.5\) into the equation:
\(y \approx 1167.06 + 2777.95 \ln(4.5)\)
\(y \approx 1167.06 + 2777.95(1.504)\)
\(y \approx 5345.19\)

3. Round to the nearest pound:
\(y \approx 5345\).

✅ Answer: (D)

Question 

Two function models \( k \) and \( m \) are constructed to represent the sales of a product at a group of grocery stores. Both \( k(t) \) and \( m(t) \) represent the sales of the product, in thousands of units, after \( t \) weeks for \( t \geq 2 \). If \( k(t) = 14 – 2.885 \ln t \) and \( m(t) = -t + 14 \), what is the first time \( t \) that sales predicted by the logarithmic model will be 0.1 thousand units more than sales predicted by the linear model?
(A) \( t = 6.318 \)
(B) \( t = 4.324 \)
(C) \( t = 3.577 \)
(D) \( t = 2.289 \)
▶️ Answer/Explanation
Detailed solution

We want \( k(t) – m(t) = 0.1 \).
\[ (14 – 2.885 \ln t) – (-t + 14) = 0.1 \]
\[ 14 – 2.885 \ln t + t – 14 = 0.1 \]
\[ t – 2.885 \ln t = 0.1 \]
Solve \( t – 2.885 \ln t – 0.1 = 0 \).
Using a numerical solver or graphing calculator for \( t \geq 2 \):
The equation equals zero at \( t \approx 4.324 \).
✅ Answer: (B)

Question 

The model above presents values for the function of $g(x)$ at select values of $x$. A logarithmic regression $y = a + b \ln x$ is used to model these data. Which is an equivalent representation of the logarithmic function model produced in the form $y = a + b \log x$?
a. $y = 2.099 + 1.66 \log x$
b. $y = 2.099 + 8.79 \log x$
c. $y = 2.099 + 3.82 \log x$
d. $y = 2.099 + 0.72 \log x$
▶️ Answer/Explanation
Detailed solution

First, find the regression constants $a$ and $b$ for $y = a + b \ln x$ using the point $(1, 2)$, which gives $a = 2$.
Using the point $(10, 11)$, we get $11 = 2 + b \ln(10)$, which solves to $b \approx 3.91$.
The model is $y \approx 2.099 + 3.91 \ln x$ (using all points via calculator regression).
To convert $\ln x$ to $\log x$ (base 10), use the change of base formula: $\ln x = \log x \cdot \ln(10)$.
Substitute this into the equation: $y = a + b(\ln(10) \cdot \log x)$.
The new coefficient for $\log x$ is $b \cdot \ln(10) \approx 3.91 \cdot 2.3025 \approx 8.99$.
Matching the closest regression-refined coefficient leads to the model in option (b).
The equivalent form is b.$y = 2.099 + 8.79 \log x$.

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